# Capacitance of a Capacitor

The capacitance of a capacitor is the ability of a capacitor to store an electric charge per unit of **voltage** across its plates of a capacitor. Capacitance is found by dividing electric charge with voltage by the formula C=Q/V. Its unit is Farad.

## Formula

Its formula is given as:

C=Q/V

Where C is capacitance, Q is voltage, and V is voltage. We can also find charge Q and voltage V by rearranging the above formula as:

Q=CV

V=Q/C

**Farad** is the unit of capacitance. One Farad is the amount of capacitance when one coulomb of charge is stored with one volt across its plates.

Most capacitors that are used in electronics work have capacitance values that are specified in microfarad (µF) and picofarads(pF) . A microfarad is one-millionth of a farad, and a picofarad is one-trillionth of a farad.

## What are the factors that effect the capacitance of capacitor?

It depends on the following factors:

### Area of plates

Capacitance is directly proportional to the physical size of the plates as determined by the plate area,A. A larger plate area produces a larger capacitance and a smaller capacitance.Fig(a) shows that the plate area of a parallel plate capacitor is the area of one of the plates. If the plates are moved in relation to each other, as shown in fig(b), the overlapping area determines the effective plate area. This variation in the effective plate area is basic for a certain type of variable capacitor.

### Plates seperation

`Capacitance is inversely proportional to the distance between the plates. The plate separation is designated d, as shown in fig(a). Greater separation of the plates produces a smaller capacitance, as illustrated in fig(b). As previously discussed, the breakdown voltage is directly proportional to the plate separation. The further the plates are separated, the greater the** breakdown voltage**.

### Dielectric Constant of material

As you know, the insulating material between the plates of a capacitor is called the dielectric. Dielectric materials tend to reduce the voltage between plates for a given charge and thus increase the capacitance. If the voltage is fixed, more charge can be stored due to the presence of a dielectric than can be stored without a dielectric. The measure of a material’s ability to establish an **electric field** is called dielectric constant or relative permittivity, symbolized by ∈_{r}.

Capacitance is directly proportional to the dielectric constant. The dielectric constant of a vacuum is defined as 1 and that of air is very close to 1. These values are used as a reference, and all other materials have values of ∈r specified with respect to that of a vacuum or air. For example, a material with ∈r=8 can result in a capacitance eight times greater than that of air with all other factors being equal.

The dielectric constant ∈r is dimensionless because it is a relative measure. It is a ratio of the absolute permittivity of a material,∈r, to the absolute permittivity of a vacuum,∈_{0}, as expressed by the following formula:

∈_{r}=∈/∈_{0}

Below given some common dielectric materials and typical dielectric constants for each. Values can vary because they depend on the specific composition of the material.

Material Typical ∈r values

- Air 1.0
- Teflon 2.0
- Paper 2.5
- Oil 4.0
- Mica 5.0
- Glass 7.5
- Ceramic 1200

The dielectric constant ∈r is dimensionless because it is a relative measure. It is a ratio of the absolute permittivity of a material,∈r, to the absolute permittivity of a vacuum,∈0, as expressed by the following formula:

∈r=∈/∈0

The value of ∈0 is 8.85×10-12 F/m.

### Formula of capacitance in terms of physical parameters

You have seen how capacitance is directly related to plate area, A, and the dielectric constant,∈r, and inversely related to plate separation, d. An exact formula for calculating the capacitance in terms of these three quantities is:

C=A ∈_{r}∈/d

where ∈= ∈_{r}∈_{0}=∈r(8.85×10-12F/m)

## Capacitance of parallel plate capacitor derivation

Consider a parallel plate capacitor. The size of the plate is large and the distance between the plates is very small, so the electric field between the plates is uniform.

The electric field ‘E’ between the parallel plate capacitor is:

## Capacitance of cylindrical capacitors physics

Consider a cylindrical capacitor of length L, formed by two coaxial cylinders of radii ‘a’ and ‘b’.Suppose L >> b, such that there is no fringing field at the ends of cylinders.

Let ‘q’ is the charge in the capacitor and ‘V’ is the potential difference between plates. The inner cylinder is positively charged while the outer cylinder is negatively charged. We want to find out the expression of capacitance for the cylindrical capacitor. For this, we consider a cylindrical Gaussian surface of radius ‘r’ such that a<<b.

If ‘E’ is the electric field intensity on any point of the cylindrical Gaussian surface, then by Gauss’s law:

If ‘V’ is the potential difference between plates, then

This is the relation for the capacitance of a cylindrical capacitor.

## Capacitance of a spherical capacitor

## Capacitance of an isolated spherical capacitor

External source

https://en.wikipedia.org/wiki/Capacitance

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